If a triangle and a convex quadrilateral are drawn on the same base and no part of the quadrilateral is outside the triangle, -Maths 9th

If a triangle and a convex quadrilateral are drawn on the same base and no part of the quadrilateral is outside the triangle, -Maths 9th
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If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th

Description : Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th

Last Answer : answer:

In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

Description : In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

Last Answer : △ABC is an equilateral triangle. ∠BAE=60 ABEC is a cyclic quadrilateral. We know, ∠BAC+∠BEC=180 =>∠BEC=180−60 =120 We know that the angles subtended by an arc on the circumference on the same side are equal. Therefore, ∠ABC=∠BDC =>∠BDC=60

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Let each side of ㎝ equilateral triangle ABC be ′a′㎝ Now, ar△OAB=21 AB OP=21 a 14=7a㎠→1 ar△OBC= BC OQ =21 a 10=5a㎠→2 ar△OAC=21 AC OR=21 a 6=3a㎠→3 ∴ar△ABC=1+2+3=7a+5a+3a=15a㎠ Also area of equilateral ... ABC=43 a2 Now, 43 a2=15a⇒a=3 15 4 3 3 =3603 =203 ㎝ Now, ar△ABC=43 (203 )2=3003 ㎠

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : NEED ANSWER

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Area of triangle =

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : This answer was deleted by our moderators...

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Last Answer : answer:

Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; -Maths 9th

Description : Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; -Maths 9th

Last Answer : answer:

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

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A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th

Description : If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th

Last Answer : Solution :-

A triangle and a parallelogram have the same base and the same area. -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. -Maths 9th

Last Answer : Let a = 26 cm, b = 28 cm, c = 30 cm ∴ s = (a + b + c)/2 = (26 + 28 + 30)/2 = 84/2 = 42 ∴ Area of triangle = root under (√s(s - a )(s - b)(s - c) = root ... cm2 Now, Area of parallelogram = Area of triangle ⇒ Base x height = 336 ⇒ 28 x height = 336 ⇒ height = 336/28 = 12 cm

The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

How many convex circles can be drawn in a triangle ?

Description : How many convex circles can be drawn in a triangle ?

Last Answer : Answer : It is possible to draw three convex circles of a triangle.

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th

Description : Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th

Last Answer : Area of a triangle = 1/2 × Base × Altitude ( height ) therefore., Area of a triangle = 1/2 × 20 cm× 10cm = 10cm ×10cm = 100 cm^2

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.

Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Description : Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Last Answer : If h is the height of the triangle, then h 2 =y 2 − 4 x​ 2 ⇒h= 4 4y 2 −x 2 ​ ​ cm ∴Area= 2 1​ ×base×h = 2 x​ 4 4y 2 −x 2 ​ ​ cm 2

The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2

The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

Description : The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

Last Answer : Given, base of an isosceles triangle =24 cm Area of isosceles triangle =192 sq.cm Area = 21​×b×h ∴192=224.h​ ∴h=16 cm Side=h2+122​=256+144​=20 cm Perimeter of triangle =2a+b =2(20)+24=64 cm

The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Description : The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Last Answer : Volume of a right prism = Area of base height. Since the base is an equilateral triangle of side 6 cm, Area of base = 3√434 x (side)2 = (3√4 62)(34 62)cm2 = 3√434 x 36 cm2 = 93-√93 cm2 ∴ Volume = (93-√93 x18) ... ) = (324 + 2 9√3 ) cm2 = (324 + 18√3 ) cm2 = (324 + 31.176) cm2 = 355.176 cm2.

The base in a right prism is an equilateral triangle of side 8 cm and the height of the prism is 10 cm. The volume of the prism is -Maths 9th

Description : The base in a right prism is an equilateral triangle of side 8 cm and the height of the prism is 10 cm. The volume of the prism is -Maths 9th

Last Answer : ⇒ Area of equilateral triangle =43 ( s i d e)2 =43 ( 8)2 =43 64 ... =3 3 2 . 5 5 4 cm3. =3 3 2 . 5 5 4 cc

The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.

There are two prisms, one has equilateral triangle as a base and the other has a regular hexagon as a base. If both the prisms have equal height -Maths 9th

Description : There are two prisms, one has equilateral triangle as a base and the other has a regular hexagon as a base. If both the prisms have equal height -Maths 9th

Last Answer : answer:

A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Description : A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Last Answer : answer:

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Description : Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Last Answer : Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Description : 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Now, In ΔACD, R and S are the mid points of CD and DA respectively. , ... , PQRS is parallelogram. PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Description : 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the ... a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.

Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.